Question

The figure below shows a $2.0 \,V$ potentiometer used for the determination of internal resistance of a $2.5 \,V$ cell. The balance point of the cell in the open circuit is $75 \,cm$. When a resistor of $ 10\,\Omega $ is used in the external circuit of the cell, the balance point shifts to $65 \,cm$ length of potentiometer wire. The internal resistance of the cell is:

• A. $ 2.5\,\Omega $
• B. $ 2.0\,\Omega $
• C. $ 1.54\,\Omega $
• D. $ 1.0\,\Omega $

Answer 1

Answer: (C)

For a potentiometer, the internal resistance $(r)$ is given by
$r=R\left(\frac{l_{1}}{l_{2}}-1\right)$
Given, $R=10 \,\Omega, l_{1}=75\, cm , l_{2}=65\, cm$
$\therefore r=10\left(\frac{75}{65}-1\right)$
$=10 \times 0.154=1.54\, \Omega$